The equilibrium constant for the given reaction has to be calculated. Concept Information: Acid ionization constant K a : Acids ionize in water. Strong acids ionize completely whereas weak acids ionize to some limited extent. The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization. The ionization of a weak acid HA can be given as follows, HA (aq) → H + (aq) + A - (aq) The equilibrium expression for the above reaction is given below. K a = [ H + ][A - ] [ HA] Where, K a is acid ionization constant, [ H + ] is concentration of hydrogen ion [ A - ] is concentration of acid anion [ HA] is concentration of the acid Autoionization of water: The equation of equilibrium for autoionization of water is, H 2 O → H + + OH - K w = [H + ][OH - ] The equilibrium expression for water at 25 o C is, [H + ][OH - ] = 1 × 10 -14 Taking negative logarithm on both sides, we get − log ( [H + ][OH - ])= -log(1 × 10 -14 ) ( − log [H + ])+(-log[OH - ])= 14 ) The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation, pH + pOH = 14, at 25 o C As pOH and pH are opposite scale, the total of both has to be equal to 14. Therefore, K w = [H + ][OH - ] =1 × 10 -14 To Calculate: The equilibrium constant for the given reaction
The equilibrium constant for the given reaction has to be calculated. Concept Information: Acid ionization constant K a : Acids ionize in water. Strong acids ionize completely whereas weak acids ionize to some limited extent. The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization. The ionization of a weak acid HA can be given as follows, HA (aq) → H + (aq) + A - (aq) The equilibrium expression for the above reaction is given below. K a = [ H + ][A - ] [ HA] Where, K a is acid ionization constant, [ H + ] is concentration of hydrogen ion [ A - ] is concentration of acid anion [ HA] is concentration of the acid Autoionization of water: The equation of equilibrium for autoionization of water is, H 2 O → H + + OH - K w = [H + ][OH - ] The equilibrium expression for water at 25 o C is, [H + ][OH - ] = 1 × 10 -14 Taking negative logarithm on both sides, we get − log ( [H + ][OH - ])= -log(1 × 10 -14 ) ( − log [H + ])+(-log[OH - ])= 14 ) The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation, pH + pOH = 14, at 25 o C As pOH and pH are opposite scale, the total of both has to be equal to 14. Therefore, K w = [H + ][OH - ] =1 × 10 -14 To Calculate: The equilibrium constant for the given reaction
Solution Summary: The author explains that the equilibrium constant for the given reaction has to be calculated. The degree to which a weak acid ionizes depends on the concentration of the acid
Identify the starting material in the following reaction. Click the "draw structure" button to launch the
drawing utility.
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[1] 0 3
C10H18
[2] CH3SCH3
H
In an equilibrium mixture of the formation of ammonia from nitrogen and hydrogen, it is found that
PNH3 = 0.147 atm, PN2 = 1.41 atm and Pн2 = 6.00 atm. Evaluate Kp and Kc at 500 °C.
2 NH3 (g) N2 (g) + 3 H₂ (g)
K₂ = (PN2)(PH2)³ = (1.41) (6.00)³ = 1.41 x 104
What alkene or alkyne yields the following products after oxidative cleavage with ozone? Click the
"draw structure" button to launch the drawing utility.
and two equivalents of CH2=O
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